Problem: Solve for $x$ and $y$ by deriving an expression for $y$ from the second equation, and substituting it back into the first equation. $\begin{align*}8x+3y &= 1 \\ 8x+6y &= -2\end{align*}$
Explanation: Begin by moving the $x$ -term in the second equation to the right side of the equation. $6y = -8x-2$ Divide both sides by $6$ to isolate $y$ $y = {-\dfrac{4}{3}x - \dfrac{1}{3}}$ Substitute this expression for $y$ in the first equation. $8x+3({-\dfrac{4}{3}x - \dfrac{1}{3}}) = 1$ $8x - 4x - 1 = 1$ Simplify by combining terms, then solve for $x$ $4x - 1 = 1$ $4x = 2$ $x = \dfrac{1}{2}$ Substitute $\dfrac{1}{2}$ for $x$ back into the top equation. $8( \dfrac{1}{2})+3y = 1$ $4+3y = 1$ $3y = -3$ $y = -1$ The solution is $\enspace x = \dfrac{1}{2}, \enspace y = -1$.